# Thread: Second Derivative Test w/ no critical numbers

1. ## Second Derivative Test w/ no critical numbers

Hello,

Here is a problem that I'm really struggling with: f(x)=2-2x-x^3. F prime of this function is -2-3x^2 and when setting to zero to find critical numbers, you end up taking the sq. root of -2/3. So there are no critical numbers.

I'm supposed to determine concavity and find inflection points. Per the second derivative test rules, F prime c must equal 0 and f double prime must exist near c. Since F prime C doesn't equal zero, as there are no critical numbers, is the graph also not concave up/down and are there also no inflection points? Am I getting this right?

I'm so confused, Thank you.

2. Originally Posted by iJenn
Hello,

Here is a problem that I'm really struggling with: f(x)=2-2x-x^3. F prime of this function is -2-3x^2 and when setting to zero to find critical numbers, you end up taking the sq. root of -2/3. So there are no critical numbers.

I'm supposed to determine concavity and find inflection points. Per the second derivative test rules, F prime c must equal 0 and f double prime must exist near c. Since F prime C doesn't equal zero, as there are no critical numbers, is the graph also not concave up/down and are there also no inflection points? Am I getting this right?

I'm so confused, Thank you.
Application of the definitions:

Concave up: f'' > 0.

Concave down: f'' < 0.

The function is a cubic and therefore has a point of inflection - the x-coordinate is therefore found by solving f'' = 0.

3. So the fact that f prime has no real critical numbers is irrelevant to the second derivative test for determining concavity?

F double prime's critical point is (0,2) and is also the point of inflection. Is this correct?

4. Originally Posted by iJenn
So the fact that f prime has no real critical numbers is irrelevant to the second derivative test for determining concavity?

F double prime's critical point is (0,2) and is also the point of inflection. Is this correct?
Yes. Yes. A simple sketch graph of f(x) = 2 - 2x - x^3 confirms all this: plot y &#x3d;2-2x-x&#x5e;3 - Wolfram|Alpha